Problem 37 - Algebra and Trigonometry

Question

In Problems \(37-44,\) find \(z w\) and \(\frac{z}{w} .\) Write each answer in polar form and in exponential form. \(z=2\left(\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}\right)\) \(w=4\left(\cos \frac{\pi}{9}+i \sin \frac{\pi}{9}\right)\)

Step-by-Step Solution

Answer

In polar form: \( zw = 8 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \), \( \frac{z}{w} = \frac{1}{2} \left( \cos \frac{\pi}{9} + i \sin \frac{\pi}{9} \right) \). In exponential form: \( zw = 8 e^{i \frac{\pi}{3}} \), \( \frac{z}{w} = \frac{1}{2} e^{i \frac{\pi}{9}} \).

Step 1 - Express in Polar Form

Both of the given complex numbers are already in polar form: \[ z = 2\bigg(\text{cos} \frac{2\,\pi}{9} + i \,\text{sin} \frac{2\,\pi }{9}\bigg) \] \[ w = 4\bigg(\text{cos} \frac{\pi}{9} + i \,\text{sin} \frac{\pi}{9}\bigg) \]

Step 2 - Find the Product \(zw\) in Polar Form

Using the product formula for complex numbers in polar form \[ z w = r_1 r_2 \bigg(\cos (\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)\bigg) \] Here, \( r_1 = 2 \), \( \theta_1 = \frac{2\pi}{9} \), \( r_2 = 4 \), and \( \theta_2 = \frac{\pi}{9} \). So, \[ zw = 2 \times 4 \bigg(\cos\bigg(\frac{2\pi}{9} + \frac{\pi}{9}\bigg) + i \sin\bigg(\frac{2\pi}{9} + \frac{\pi}{9}\bigg)\bigg) \] This simplifies to: \[ zw = 8 \bigg(\cos \frac{3\pi}{9} + i \sin \frac{3\pi}{9}\bigg) \] Therefore, \[ zw = 8 \bigg(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\bigg) \]

Step 3 - Express \(zw\) in Exponential Form

Convert the result to exponential form using Euler's formula: \[ zw = 8 e^{i \frac{\pi}{3}} \]

Step 4 - Find the Quotient \(\frac{z}{w}\) in Polar Form

Using the quotient formula for complex numbers in polar form \[ \frac{z}{w} = \frac{r_1}{r_2} \bigg(\cos (\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)\bigg) \] Here, \( r_1 = 2 \), \( \theta_1 = \frac{2\pi}{9} \), \( r_2 = 4 \), and \( \theta_2 = \frac{\pi}{9} \). \ So, \[ \frac{z}{w} = \frac{2}{4} \bigg(\cos\bigg(\frac{2\pi}{9} - \frac{\pi}{9}\bigg) + i \sin\bigg(\frac{2\pi}{9} - \frac{\pi}{9}\bigg)\bigg) \] This simplifies to \[ \frac{z}{w} = \frac{1}{2} \bigg(\cos \frac{\pi}{9} + i \sin \frac{\pi}{9}\bigg) \]

Step 5 - Express \(\frac{z}{w}\) in Exponential Form

Convert the result to exponential form: \[ \frac{z}{w} = \frac{1}{2} e^{i \frac{\pi}{9}} \]

Key Concepts

polar form

Understanding the polar form of complex numbers is crucial. It represents a complex number as a combination of a magnitude (radius) and an angle. We write it as \[ r (\text{cos} \theta + i \text{sin} \theta) \].

• Magnitude, \( r \): The distance of the complex number from the origin in the complex plane.
• Angle, \( \theta \): The angle formed with the positive real axis.

In our exercise, the complex numbers are given directly in polar form:

\[ z = 2\bigg(\text{cos} \frac{2\pi}{9} + i \text{sin} \frac{2\pi }{9}\bigg)\] \[ w = 4\bigg(\text{cos} \frac{\pi}{9} + i \text{sin} \frac{\pi}{9}\bigg)\] Each number is expressed as a multiplication of its magnitude and a trigonometric expression of its angle.

exponential form

The exponential form of complex numbers provides an alternative way to represent them using Euler's formula:

\[ e^{i\theta} = \text{cos}\theta + i \text{sin}\theta \]This form is compact and convenient, especially for multiplication and division. Any complex number can be written as
\[ z = re^{i\theta} \]In our case, for the product and quotient found in the exercise, we convert these results from polar to exponential form:

• \[ zw = 8e^{i \frac{\pi}{3}} \]
• \[ \frac{z}{w} = \frac{1}{2} e^{i \frac{\pi}{9}} \]

This representation is especially useful for understanding the behavior of complex numbers in different operations.

product of complex numbers

Finding the product of two complex numbers in polar form is straightforward. The magnitudes multiply, and the angles add:
\[ z w = r_1 r_2 \bigg(\cos (\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)\bigg) \]In our given problem, we have:
\( z = 2 \text{\bigg(\cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9}\bigg)} \)\( w = 4 \text{\bigg(\cos \frac{\pi}{9} + i \sin \frac{\pi}{9} \bigg)} \)Multiplying these, we get:
\( r_1 = 2 \)\( \theta_1 = \frac{2\pi}{9} \)\( r_2 = 4 \)\( \theta_2 = \frac{\pi}{9} \)This leads to:
\( zw = 2 \times 4 \bigg(\cos\bigg(\frac{2\pi}{9} + \frac{\pi}{9}\bigg) + i \sin\bigg(\frac{2\pi}{9} + \frac{\pi}{9}\bigg)\bigg) \)
Simplifying the angles gives us:
\( zw = 8 \bigg(\cos \frac{3\pi}{9} + i \sin \frac{3\pi}{9}\bigg)\) which equals \( zw = 8 \bigg(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\bigg) \)

quotient of complex numbers

To find the quotient of two complex numbers in polar form, the magnitudes divide, and the angles subtract:
\[ \frac{z}{w} = \frac{r_1}{r_2} \bigg(\cos (\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)\bigg) \]For the given numbers:
\( z = 2 \text{\bigg(\cos \frac{2\pi}{9} + i \sin \frac{2\pi}{9} \bigg)} \)\( w = 4 \text{\bigg(\cos \frac{\pi}{9} + i \sin \frac{\pi}{9} \bigg)} \)We have:
\( r_1 = 2 \)\( \theta_1 = \frac{2\pi}{9} \)\( r_2 = 4 \)\( \theta_2 = \frac{\pi}{9} \)Taking the quotient, we get:
\( \frac{2}{4} \bigg(\cos\bigg(\frac{2\pi}{9} - \frac{\pi}{9}\bigg) + i \sin\bigg(\frac{2\pi}{9} - \frac{\pi}{9}\bigg)\bigg) \)
Simplifying the angles gives us:\( \frac{z}{w} = \frac{1}{2} \bigg(\cos \frac{\pi}{9} + i \sin \frac{\pi}{9}\bigg) \)